Question

Two infinite planes each with uniform surface charge density $+\sigma C/m^2$ are kept in such a way that the angle between them is ${30}^o$. The electric field in the region shown between them is given by:

• A. $\frac{\sigma }{2{ϵ}_0}\left[\right(1-\frac{\sqrt{3}}{2})\hat{y}-\frac{1}{2}\hat{x}]$
• B. $\frac{\sigma }{2{ϵ}_0}\left[\right(1+\frac{\sqrt{3}}{2})\hat{y}-\frac{1}{2}\hat{x}]$
• C. $\frac{\sigma }{2{ϵ}_0}\left[\right(1-\frac{\sqrt{3}}{2})\hat{y}+\frac{1}{2}\hat{x}]$
• D. $\frac{\sigma }{2{ϵ}_0}\left[\right(1+\frac{\sqrt{3}}{2})\hat{y}+\frac{1}{2}\hat{x}]$

Answer 1

The correct option is A $\frac{\sigma }{2{ϵ}_0}\left[\right(1-\frac{\sqrt{3}}{2})\hat{y}-\frac{1}{2}\hat{x}]$

$$\begin{array}{cc}& \text{Field due to single plate}=\frac{\sigma }{2{ϵ}_o}=\left[\overrightarrow{E_1}\right]=\left[\overrightarrow{E_2}\right]\\ & \text{Net electric field}\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}\\ & =\frac{\sigma }{2{ϵ}_0}\cos {30}^0(-\hat{j})+\frac{\sigma }{2{ϵ}_0}\sin {30}^0(-\hat{i})+\frac{\sigma }{2{ϵ}_0}\left(\hat{j}\right)\\ & =\frac{\sigma }{2{ϵ}_0}(1-\frac{\sqrt{3}}{2})\left(\hat{j}\right)-\frac{\sigma }{4{ϵ}_0}\left(\hat{i}\right)\\ & =\frac{\sigma }{2{ϵ}_0}[(1-\frac{\sqrt{3}}{2})\hat{y}-\frac{1}{2}\hat{x}]\end{array}$$